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Optimal Bit and Power Loading for Amplify-and-Forward Cooperative OFDM Systems

Optimal Bit and Power Loading for Amplify-and-Forward Cooperative OFDM Systems Osama Amin,Student Member,IEEE,and Murat Uysal,Senior Member,IEEE

Abstract—In this paper,we investigate bit and power allocation strategies for an orthogonal frequency division multiplexing (OFDM)cooperative network over frequency-selective fading channels.We assume amplify-and-forward relaying and consider the bit error rate(BER)performance as our performance measure.Aiming to optimize the BER under total power con-straint and for a given average data rate,we propose three adaptive algorithms;optimal power loading(OPL),optimal bit loading(OBL),and optimal joint bit and power loading(OBPL). Our Monte Carlo simulation results demonstrate performance gains through adaptive bit and power loading over conventional non-adaptive systems as well as currently available adaptive cooperative scheme in the literature.The impact of practical issues on the performance of proposed adaptive schemes such as imperfect channel estimation and limited feedback is further discussed.

Index Terms—OFDM,power allocation,bit allocation, amplify-and-forward relaying,cooperative network.

I.I NTRODUCTION

E XPLOITING the broadcasting nature of wireless trans-

mission,cooperative diversity forms a virtual antenna array among the single-antenna nodes and extracts the spatial diversity advantages in a distributed setting[1],[2].Most of the existing work in the area of cooperative diversity is based on the assumption of open-loop implementation where the destination node and(possibly)relay nodes have channel knowledge via channel estimation,while transmitting nodes have no knowledge of the channel.Such open-loop designs are favourable in time-varying channels where feedback of channel estimates becomes problematic.However,particularly for?xed wireless access applications,reliable feedback is possible and available channel state information(CSI)at the transmitting nodes can be used to design adaptive transmission schemes for performance improvement.

Adaptive transmission has been extensively studied in the context of both single-carrier and multi-carrier point-to-point links,see e.g.,[3]–[8]and the references therein.Adaptive transmission has been recently applied to cooperative diversity

Manuscript received August14,2009;revised April14,2010,August30, 2010,and October7,2010;accepted October7,2010.The associate editor coordinating the review of this paper and approving it for publication was S. Bhashyam.

This paper was presented in part at the11th Canadian Workshop on Information Theory(CWIT09),Ottawa,Ontario,Canada,May2009.

O.Amin is with the Department of Electrical and Computer Engi-neering,University of Waterloo,Waterloo,ON,N2L3G1,Canada(e-mail: oamin@uwaterloo.ca).The work of O.Amin is supported by the Egypian Higher Education Ministry.

M.Uysal is with the Faculty of Engineering,¨Ozye?g in University,34662, Istanbul,Turkey(e-mail:murat.uysal@https://www.sodocs.net/doc/1717604535.html,.tr).

Digital Object Identi?er10.1109/TWC.2011.011111.091233systems.In[9],Ahmed et al.assume a single relay sys-tem with amplify-and-forward(AF)relaying over frequency-?at Rayleigh fading channels.They propose power control schemes to minimize the outage probability for?xed rate transmission.In[10],Ahmed and Aazhang design rate and power control algorithms for decode-and-forward(DF)relay-ing.A theoretical study for the achievable rates in adaptive DF and compress-and-forward relaying is further introduced in[11].In[12],[13]Hammerstom and Wittneben consider frequency-selective channels and propose optimum power control to maximize the instantaneous rate for an uncoded two-hop OFDM scheme.Unlike[12],[13]which adopts AF relaying,Ying et al.[14]consider DF relaying and investigate adaptive OFDM cooperative systems with orthogonal AF cooperation protocol[2](also known as receive diversity (RD)protocol[15]).In[16],Gui and Cimini also consider RD protocol and present bit and power loading algorithms for OFDM DF cooperative systems with sub-carrier selection to minimize the total transmission power under a?xed rate assumption.In another work which builds on RD protocol [17],[18],Ma et al.propose a sub-optimal bit and power loading algorithm for OFDM cooperative system to maximize the throughput under individual power constraints and a target link error rate assuming AF and selective DF relaying.In[19], the same authors study bit and power loading algorithms to minimize the transmit power consumption for AF and selective DF modes at a target throughput.In[20],Hajiaghayi et al. address power loading for an OFDM AF system with RD protocol.They formulate two problems;one aims to maximize the system capacity and the other aims to minimize the bit error rate(BER).

In most of the existing literature on adaptive OFDM[12]–[14],[16]–[20],objective function is chosen as either to maximize the throughput or minimize the power consumption. An exception is[20]which addresses power loading for adap-tive OFDM to minimize BER performance in a single-relay scenario.However,their solution is sub-optimal since they work under the high signal-to-noise ratio(SNR)assumption and consider identical subcarrier power at source and relay nodes.In this paper,we aim to optimize BER performance and propose three adaptive bit or/and power loading algorithms for single-relay OFDM system with AF relaying and RD protocol.The?rst algorithm computes the optimal source and relay power loading coef?cients under total power constraint and?xed subcarriers’rate.The second algorithm computes the optimal bit loading coef?cients under?xed average trans-mission rate and equal power loading.The third algorithm computes the joint optimal power and bit loading coef?cients under total power constraint and?xed average transmission

1536-1276/11$25.00c?2011IEEE

rate.Through Monte-Carlo simulations,we demonstrate that our proposed schemes achieve full diversity and outperform conventional schemes with equal power loading as well as precoded systems.We further investigate the effect of practical considerations such as imperfect CSI and quantization on the performance.

The rest of the paper is organized as follows:The signal model for cooperative OFDM system under consideration is described in Section II.Optimization problem to minimize BER is ?rst formulated in Section III and then correspond-ing power and/or bit loading algorithms are presented.The simulated performance of proposed adaptive schemes is pre-sented in Section IV .And ?nally Section V summarizes and concludes the paper.

Notation :Bold upper-case letters denote matrices and bold lower-case letters denote vectors.E and ∣.∣denote respectively the expectation and the absolute value.(.)T and (.)H denotes transpose and conjugate transpose (i.e.,Hermitian)operations,respectively.Q (.)represents the Gaussian Q function.Q represents the N ×N fast Fourier transform (FFT)matrix,i.e.Q (p,q )=(1/√N )exp(?j 2π(p ?1)(q ?1)/N ),p =0,1,...,N ?1,q =0,1,...,N ?1.

II.T RANSMISSION M ODEL

We consider a cooperative OFDM system with single relay.Source,relay,and destination nodes are equipped with single transmit/receive antennas and operate in half-duplex mode.The nodes are assumed to be located in a two-dimensional plane where d SD ,d SR and d RD denote the distances of source-to-destination (S →D),source-to-relay (S →R),and relay-to-destination (R →D)links,respectively (see Fig.1).In Fig.1,θis the angle between lines representing S →R and R →D links.

To explicitly take into account the effect of relay location,we consider both long-term path loss and short-term frequency-selective Rayleigh fading.The path loss is proportional to d αwhere d is the distance between nodes and αis the path loss exponent.By normalizing the path loss terms with respect to the direct S →D link,the so-called geometrical gains can be de ?ned

as G SR =(d SD /d SR )αand G RD =(d SD /d RD )α

[15].These are related through the cosines law by G 2/αSR +G 2/αRD ?2G 1/αSR G 1/αRD cos θ=G 2/αSR G 2/α

RD .The frequency-selective fading channels are modeled as FIR (?nite impulse response)?lters with order of L SD ,L SR and L RD .They are represented by h SD =[?SD (0),...,?SD (L SD )]T ,h SR =[?SR (0),...,?SR (L SR )]T and h RD =

[?RD (0),...,?RD (L RD )]T

for S →D,S →R,and R →D links,respectively.The entries of h SD ,h SR ,and h RD are assumed to be zero mean,complex Gaussian distribution with their variance equal to 1/(L SD +1),1/(L SR +1)and 1/(L RD +1),respectively.The channels are assumed to remain constant over a block of OFDM symbols and change from one block to another independently.

Here,we assume RD cooperation protocol where the source and the relay nodes transmit in orthogonal transmission phases.In the ?rst transmission phase,a bit-stream is fed into serial-to-parallel converter which maps them into modulation

Fig.1.

Cooperative system model.

symbols chosen from either M-PSK or M-QAM constellations.Each k subcarrier symbol carries b k bits based on the em-ployed bit loading algorithm (which will be later introduced).Before passing through inverse FFT (IFFT),the power of each subcarrier symbol is adjusted based on the employed power loading algorithm (which will be later introduced).To prevent inter-block interference,a cyclic pre ?x (CP)is inserted be-tween OFDM symbols with L CP ≥max(L SR ,L RD ,L SD ).Both relay and destination nodes receive the transmitted OFDM symbol.After removing CP and converting the OFDM symbol into parallel subcarrier symbols through FFT,the relay node scales the subcarriers power.Then it feeds the subcarrier symbols to IFFT and adds CP.In the second transmission phase,the relay node forwards the resulting signal to the desti-nation while the source node remains silent.At the destination,both OFDM symbols received during the broadcasting and relaying phases are fed to maximum likelihood (ML)detector after removing CP and passing through FFT.Block diagrams of source,relay,and destination nodes are provided in Fig.2.Let the subcarrier signal for the k th carrier be denoted as x (k ),k =1,2,...N where N is the number of subcarriers.The received signals at the relay and the destination nodes during the broadcasting phase are given by

r D 1(k )=√

E k,S exp (j?k,S )D SD (k,k )x (k )+v D 1(k )(1)

r R (k )=√

E k,S G SR exp (j?k,R )D SR (k,k )x (k )+v R (k )(2)

where E k,S and ?k,S denote,respectively,the adjustable power and phase terms for k th subcarrier.In the above v D 1(k ),and v R (k )are the FFT of additive white Gaus-sian noise (AWGN)terms n D 1(k )and n R (k ).D SD (k,k )and D SR (k,k )are the k th diagonal elements in the di-agonal channel matrices D SD and D SR .They are de ?ned as D SD =diag(H SD (0),H SD (1),???H SD (N ?1))and D SR =diag(H SR (0),H SR (1),???H SR (N ?1))where H SD (k )=∑N ?1l =1?SD (l )exp(?j 2πlk/N )and H SR (k )=∑N ?1

l =1?SR (l )exp(?j 2πlk/N ).For the scaling at the relay node,we adopt a slightly modi ?ed version of so-called instan-taneous power scaling (IPS)[21],[22]which assumes perfect CSI.In IPS,scaling term is given by

v R (k )

{∣r R (k )∣2}

=E k,S G SR ∣D SR (k,k )∣2

+N 0.(3)Here,instead we replace E k,S with the average value E .

Such a modi ?cation does not signi ?cantly affect the overall performance,but simpli ?es the ensuing optimization problem.After scaling the received signal,the relay node ampli ?es

(a)Source

(b)Relay

(c)Destination

Fig.2.Block diagrams of source,relay,and destination nodes.

the k th subcarrier with power E k,R and adjusts the phase by adding?k,R.Then it feeds the subcarriers to IFFT and adds CP before it forwards the resulting signal to destination node. The destination node removes CP and converts the received OFDM symbol into subcarrier symbols.The k th subcarrier signal is given by

r D

2(k)=

√

G SR G RD E k,S E k,R

√

EG SR∣D SR(k,k)∣2+N0

exp(j(?k,S+?k,R))

×D SR(k,k)D RD(k,k)x(k)+v′D

(k)(4)

In(4),v′D

2(k)represents the effective noise term and is

given by

v′D

2(k)=

√

G RD E k,R D RD(k,k)exp(j?k,R)

√

EG SR∣D SR(k,k)∣2+N0

v R(k)+v D

(k)

(5)

where v D

2(k)is the FFT of AWGN term at the

destination during the relaying phase.Effective noise is conditionally Gaussian with zero mean and variance of?=√

1+E k,R G RD∣D RD(k,k)∣2/(EG SR∣D SR(k,k)∣2+N0).Normalizing(5)with?,we have

?r D

(k)=

√

E k,S E k,R D k,2exp(j(?k,S+?k,R))

√

E k,R D k,R+D k,0

x(k)+?v D

(k)

(6) where?v D

(k)is the output noise with variance N0.In(6),D k,0,D k,1,D k,2,and D k,R are de?ned as D k,0=EG SR∣D SR(k,k)∣2+N0,D k,1= D SD(k,k),D k,2=

√

G SR G RD D SR(k,k)D RD(k,k) and D k,R=G RD∣D RD(k,k)∣2.Based on the received signals given by(1)and(6)and assuming perfect CSI,the destination node performs ML detection using the metric

?x(k)=arg min

x(k)

{

D1(k)?

√

E k,S exp(j?k,S)D k,1x(k)

2 +

?r D2(k)?

√

E k,S E k,R exp(j(?k,S+?k,R))D k,2

√

E k,R D k,R+D k,0

x(k)

}

(7)

III.A DAPTIVE L OADING A LGORITHMS FOR BER

O PTIMIZATION

In this section,we propose three adaptive bit and/or power loading algorithms to minimize the BER.Based on the avail-ability of instantaneous CSI,approximate BER expressions for M-PSK and rectangular M-QAM are given by[23]

P≈

∑

k=1

c k Q(

√

a kεk)(8) where c k,a k andεk are de?ned as

εk=E k,S∣D k,1∣2+

E k,S E k,R∣D k,2∣2

(E k,R D k,R+D k,0)

(9)

c k=

(M k?1)

∑

k=1

log2M k

,for M?PSK

∑

k=1

log2M k

(√

M k?1

√

M k

)

,for M?QAM

(10)

a k=

{

sin2

(

M k

)

,for M?PSK

N0(M k?1)

,for M?QAM

(11) where M k is the constellation size for the k th subcarrier. A.Optimal Power Loading(OPL)

In this subsection,we aim to?nd OPL coef?cients for source and relay subcarriers to minimize the BER under total power constraint and?xed subcarrier rate(i.e.,?xed modulation scheme for all subcarriers).Total power constraint dictates(1/N)

∑N

k=1

(E k,S+E k,R)=2E.On the other hand,under?xed subcarrier rate,we have M k=M which yields constant values c k=c and a k=a for all subcarriers. Let us de?ne E k,S=[E1,S E2,S???E N,S]T and E k,R=[E1,R E2,R???E N,R]T as the vectors repre-senting source and relay power loading coef?cients.Therefore, the optimization problem can be expressed as

min

E k,S,E k,R

∑

k=1

√

aεk)(12)

subject to the constraint of

1N N ∑

k =1

(E k,S +E k,R )=2E,E k,S >0,E k,R ≥0(13)

Since Q (.)is convex,objective function in (12)is also convex as proved in the appendix.Therefore,its solution will provide global optimum results.

Factoring the constraint into the objective function,we formulate Lagrangian problem as

Ψ=c

N ∑k =1Q (√aεk )+λ(1

N ∑k =1

(E k,S +E k,R )?2E )?

N ∑

k =1

μk E k,S ?

N ∑k =1

ηk E k,R

(14)

where λis the Lagrange multiplier.Eq.(13)can be rewritten as

?g Ψ=0

(15)

where we de ?ne g as

g =[E 1,S ,???,E N,S ,E 1,R ,???,E N,R ,μ1,???μN ,

η1,???ηN ,λ,?1,S ,???,?N,S ,?1,R ,???,?N,R ]

(16)and ?g is the gradient operator with respect to elements of g .Since Ψis independent of ?k,S and ?k,R ,they will not affect the optimization.Therefore,we set ?k,S =?k,R =0.Karush-Kuhn-Tucker (KKT)conditions for the optimization problem at hand can be then written as [24]

?Ψ?E k,S =?c √a 8πεk exp (?aεk 2)?εk

?E k,S +

λN

?μk =0(17)?Ψ?E k,R =?c √a 8πεk exp (?aεk 2)?εk

?E k,R +

λN

?ηk =0(18)

N ∑

k =1

(E k,S +E k,R )?2E =0(19)μk E k,S =0(20)ηk E k,R =0(21)E k,R ≥0,E k,S >0(22)ηk ≥0,μk ≥0,λ≥0

(23)

Optimum values of power loading coef ?cients E k,S ,E k,R can be then obtained by simultaneously solving (17)-(23).Eqs.(22)and (23)have the following four possible solutions:

μk =0,ηk =0(24)μk =0,E k,R =0(25)ηk =0,E k,S =0(26)E k,S =0,E k,R =0

(27)

The third and fourth solutions in (26)and (27)are not feasible and can be ignored under our assumption of ?xed subcarrier rate.The second solution (25)means no coopera-tion for the k th subcarrier and is adopted only for unreliable subcarriers (which will be elaborated later).On the other hand,

solution (24)represents the cooperation case which is adopted

for reliable subcarriers.Inserting μk =0,ηk =0in (17)and (18),we can readily ?nd out ?εk /?E k,S =?εk /?E k,R .Us-ing this relation and after some mathematical mainpulations,we can express E k,S in terms of E k,R as

E k,S =D k,R D k,0(

∣D k,1∣2

D k,R ∣D k,2∣

2+1)E 2

k,R +(2D k,R ∣D k,1∣2∣D k,2∣2+1)

E k,R +D k,0∣D k,1∣

∣D k,2∣2

(28)then we can rewrite (14)in terms of E k,R as Ψ=c

N ∑k =1

Q (√

aεk )

+λ

(

1N N ∑k =0

(D k,R D k,0

γk E 2

k,R +2γk E k,R +αk )?2E

)(29)

where εk is expressed in terms of E k,R as

εk =∣D k,2∣2

D k,0

(γk E k,R +αk )2

(30)

Here,αk and γk are de ?ned as

αk =D k,0∣D k,1∣2

∣D k,2∣

(31)γk =1+D k,R ∣D k,1∣2/∣D k,2∣

(32)

Solving ?Ψ/?E k,R =0yields

f (E k,R )=exp (?a ∣D k,2∣2

2D k,0

(γk E k,R +αk )

)

?λ(√8π(D k,R E k,R +D k,0)

cN ∣D k,2∣√aD k,0

)

(33)

From (33),it can be found out that some relay subcarriers will have negative power values if the following condition given as

exp (?a ∣D k,2∣2

2D k,0α2

k )<λ

(√

8πD k,0cN ∣D k,2∣√a )(34)is satis ?ed.These basically correspond to unreliable relay

subchannels.Since negative power is physically meaningless,the power of such subcarriers should be forced to zero.This indicates that the optimum solution will yield the non-cooperative case for the unreliable subcarriers.Replacing μk =0,E k,R =0in (17)and (18)and imposing KKT conditions,we have

E k,S =1a ∣D k,1∣2

W ??(∣D k,1∣2

caN 2√2πλ)2?

?(35)where W in (35)is the Lambert function,i.e.,the inverse

function of f (x )=x exp (x )[25].Based on the above derivation steps,we can summarize the proposed algorithm as in Algorithm 1.

As an example,we illustrate power distribution among subcarriers for given CSIs in Fig.3.The curves labelled by

Algorithm 1OPL Algorithm

1:Choose an initial interval for λas [λa ,λb ].

2:For each subcarrier,check if (33)is satis ?ed.If it is valid,set the subcarriers’power as E k,R =0and E k,S =1a ∣D k,1

∣2W ((∣D k,1

∣2

caN 2√2πλ

)2).If (33)is not satis ?ed,compute E k,R and E k,S by solving

the following equations,respectively,exp (

?a ∣D k,2∣22D k,0(γk E k,R +αk )

)?λ(√

8πcN ∣D k,2∣√aD k,0)(D k,R E k,R +D k,0)=0,E k,S =D k,R D k,0(∣D k,1∣2D k,R ∣D k,2∣2

+1)E 2

k,R +(2D k,R ∣D k,1∣2∣D k,2

∣2+1)E k,R +D k,0∣D k,1∣2

∣D k,2∣2.3:Compute the power constraint function given as

C f (E S ,E R )=1N N ∑k =1

(E k,S +E k,R )?2E

Update the [λa ,λb ]interval and repeat from step 1until

convergence,i.e.,the C f reaches zero.

Fig.3.

Power and loading coef ?cients for a given channel realization.

D SD ,D SR ,and D RD denote fading channel realizations.The curves labelled by

E S and E R denote the power loading coef ?cients for source and relay assuming N =32.It can be observed that OPL algorithm effectively equalizes bad sub-channels assigning more power onto them.

B.Optimal Bit Loading (OBL)

In this subsection,we aim to ?nd OBL coef ?cients assum-ing equal power loading and ?xed average transmission rate.Under these assumptions,we have E k,S =E k,R =E ?k

and ∑N k =1b k =NB where b k =log 2M k denotes the bit value assigned to the k th subcarrier bit loading coef ?cient.Therefore,the optimization problem is given by

min

N ∑k =1

c k Q (√

a k εk )

(36)

subject to the following constraints

N ∑k =1

b k =NB,0≤b k ≤NB,E k,S =E k,R =E ?k (37)

The optimization problem of ?nding b =[b 1,b 2,...b N ]

is classi ?ed as an integer optimization problem of separable objective functions.The optimal solution can be found using the dynamic programming approach [26]where the objective function in (36)is simpli ?ed by dividing it into simpler sub-problems which are then solved recursively.In our case,we de ?ne the recursive functions associated with sub-problems as

(k )

(p )=min

b j

k ∑j =1

f j (b j )(38)

subject to ∑k

j =1b j

=p .Here,f j (b j )=c j Q (√a j εj ),b j is a non-negative integer j =1,2,???,k ,k =1,2,???,N and p =0,1,???,NB .Starting with the initial condition f (1)(p )=f 1(p ),p =0,1,???,NB we recursively need to compute f (k )(p )which are functions of c p and a p ,c.f.(10)and (11).At k =N and p =NB ,we obtain f (N )(NB )which yields the OBL coef ?cients.The proposed algorithm can be summarized as in Algorithm 2.As an example,we illustrate bit loading values for given CSIs in Fig.3.As expected,OBL algorithm loads more bits to reliable channels.Algorithm 2OBL Algorithm

1:Initiate k =1.Calculate f (1)(p )as f (1)(p )=f 1(p )=c p Q (√a p ε1

)2:De ?ne b k (p )=p 3:k =k +1

For p =0,1,???,NB and l =0,1,???,p ,compute f (k ?1)(p ?l )+f k (l ).Let l min denote the value of l which minimizes f (k ?1)(p ?l )+f k (l ).Set f (k )(p )=f (k ?1)(p ?l min )+f k (l min )and b k (p )=l min .5:

If k

For k =N and p =NB ,compute f (N ?1)(NB ?l )+f N (l ).Let b N (NB )denote the value of l which mini-mizes f (N ?1)(NB ?l )+f N (l ).

Compute OBL coef ?cients b as follows,b =[b opt (1),b opt (2),...b opt (N )]b opt (N )=b q (NB )where q =N and

q =q ?1b.p opt =N ?∑N j =q ?1b opt (j )c.b opt (q )=b q (p opt )

If q >1go to step 7.a otherwise stop.

C.Optimal Bit and Power Loading (OBPL)

In the previous two problems,we have ?xed either subcar-rier power or subcarrier rate and computed the optimal value for the other parameter.In this section,we present a joint OBPL scheme which simultaneously optimizes subcarrier rate and power to minimize BER.This problem can be expressed

min

E S ,E R ,b

γ(E S ,E R ,b )=

N ∑k =1

c k Q (√

a k εk )

(39)

subject to the following constraints

1N N ∑

k =1(E k,S +E k,R )=2E,E k,S >0,E k,R ≥0,

N ∑k =1

b k =B,0≤b k ≤NB

(40)

The current problem is a mixed integer nonlinear convex optimization problem [27].Here,we use Generalized Benders Decomposition (GBD)method [27]which converts the joint optimization problem into upper and lower bound optimization problems.The upper bound problem is a nonlinear opti-mization problem for the computation of the power loading coef ?cients.On the other hand,the lower bound problem is an integer optimization problem for the computation of the bit loading coef ?cients.The joint optimal solution can be obtained by iteratively solving the two problems each of which uses the others output until convergence.The proposed OPBL algorithm based on GBD method can be summarized as in Algorithm 3.

Algorithm 3OBPL Algorithm

For q =1initiate b =b (q )and set α0=?∞,LB 0=?∞and UB 0=∞.

2:For a given bit loading coef ?cients vector b (q ),?nd OPL

coef ?cients E (q )S ,E (q )

R and the Lagrange multiplier λ(q )by solving the upper bound optimization problem using the OPL algorithm (introduced in Section III-a).3:Set UB (q )

=min {UB (q ?1),γ(E (q )S ,E (q )R ,b (q ))}

If UB (q )=γ(E (q )S ,E (q )

R ,b (q )),set (E S,opt ,E R,opt )=(E (q )S ,E (q )R ).

5:De ?ne Lagrangian function V (.)as V (E S ,E R ,b ,λ)=γ(E S ,E R ,b )+λ(1

N ∑k =1(E k,S +E k,R )?2E )Using the OPL coef ?cients (E (q )S ,

E (q )

R )obtained in step

2,update the bit loading coef ?cients b through solving the lower bound integer optimization problem

min b (q +1)V (E (q )S ,E (q )R ,b (q ),λ(q )

)+?T M V (

E (q )S ,E (q )R ,b (q ),λ

(q ))(2b ?2b (q )

)subject to ∑N k =1b (q )

k =NB and α(q )≥α

(q ?1)

,where α(q )=V (E (q )S ,E (q )

R ,b (q ),λ(q )

)+?T M V (

E (q )S ,E (q )R ,b (q ),λ(q ))(2b (q +1)?2b (q )

).(see Algorithm 4for performing step (5))6:Set LB (q )=α(q )

7:If LB (q )≥UB (q ),stop and the optimal solution will be given as (E S,opt ,E R,opt ,b opt )=(

E (q )S ,E (q )R ,b

(q )

).Otherwise,set q =q +1and go to step 2.

Algorithm 4OBPL Sub-algorithm

Let y i (p )=??M i V (E (q )i,S ,E (q )i,R ,b (q )i ,λ

(q ))(2p ?2b (q )i

)1:Initiate k =1.For p =0,1,???,NB ,compute y (1)(p )=V (

E (q )1,S ,E (q )1,R ,b (q )

1,λ(q ))+y 1(p ).2:De ?ne b k (p )=p 3:Set k =k +1

4:For p =0,1,???,NB and l =0,1,???,p ,compute y (k ?1)(p ?l )+y k (l ).Let l min denote the value of l which minimizes y (k ?1)(p ?l )+y k (l ).Set b k (p )=l min and y (k )(p )=y (k ?1)(p ?l min )+y k (l min ).

5:If k

6:For k =N and p =NB ,compute y (N ?1)(NB ?l )+y N (l ).Let b N (NB )denote the value of l which mini-mizes y (N ?1)(NB ?l )+y N (l )subject to α(q )≥α(q ?1).7:Compute OBL coef ?cients b as follows,b =[b opt (1),b opt (2),...b opt (N )],b opt (N )=b q (NB )where q =N and

a.q =q ?1

b.p opt =N ?∑N j =q ?1b opt (j )

c.b opt (q )=b q (p opt )

If q >1go to step 7.a otherwise stop.

IV.S IMULATION R ESULTS

In this section,we investigate the BER performance of proposed OPL,OBL,and OBPL algorithms through Monte Carlo simulations.We assume α=2,θ=πand N =32.The channel lengths for the three links are assumed to be equal to 2,i.e.,L SR =L RD =L SD =L =1,SNR is de ?ned to be E/N 0,where E is the average subcarrier power.Example 1(Performance of OBPL)

In this simulation example,we study the performance of the OBPL algorithm with perfect CSI assuming an average transmission rate of 2and 4bits/subcarrier,respectively,in Figs 4and 5.We consider G SR /G RD =30dB,?30dB,and 0dB.These respectively correspond to cases where the relay is close to the source,close to the destination,and at the midpoint between the source and the destination.The performance of non-adaptive scheme with equal bit and power loading (EBPL)is included as a benchmark.For comparative observations on the diversity order,the performances of co-located multi-antenna single-input single-output (SISO)systems with 2,3and 4antennas assuming maximal ratio combining (MRC)are included as further benchmarks.

As observed in Figs.4and 5,EBPL scheme achieves only a diversity of two.This diversity gain results from the spatial diversity for the single-relay scenario under consideration.Therefore,EBPL system is not able to extract the underlying multipath diversity.On the other hand,the proposed OBPL scheme is able to extract a diversity order of 2(L +1)=4and signi ?cantly outperforms EBPL scheme.For example,in Fig.2,at BER=10?3,OBPL outperforms EBPL by 3.5dB assuming G SR /G RD =0dB.This climbs up to 5.8dB for G SR /G RD =?30dB.It is also interesting to note that each scheme attains its best performance at different locations.EBPL attains its best performance when the relay is at the midpoint (i.e.,G SR /G RD =0dB).On the other hand,the performance of OBPL gets better when the relay is placed

Fig.4.Performance

comparison of EBPL and OBPL schemes for different relay locations with average transmission of 2bits/subcarrier.Fig.5.Performance comparison of EBPL and OBPL schemes for different relay locations with average transmission of 4bits/subcarrier.

close to the destination (i.e.,G SR /G RD =?30dB).Similar observations hold for Fig.3where an average transmission rate of 4bits/subcarrier is assumed,where at BER=10?3,OBPL outperforms EBPL by 3dB assuming G SR /G RD =0dB.Also the improvement reaches its maximum at G SR /G RD =?30dB where 5dB is achieved.From these results,we conclude that dealing with higher rates reduces performance improvement.

Example 2(OPBL versus other Schemes)

In this example,we ?rst compare the performance of OBPL to OPL and OBL schemes.Through this comparison,we are particularly interested in ?nding out whether bit or power loading is more rewarding in performance optimization.In our simulations,we assume perfect CSI,an average transmission rate of 2bits/subcarrier and G SR /G RD =?30dB.From Fig.6,we observe that,at BER=10?4,performance gap between OBPL and OBL is 3.6dB.On the other hand,performance gap between OBPL and OPL reduces to 1.1dB.Therefore,it can be concluded that power loading is more dominant in performance

optimization.

Fig.6.Performance comparison of OBPL,OPL,OBL algorithms and competing schemes.

For comparison with existing systems in the literature,we also include the performance of a precoded cooperative OFDM [28]in Fig.6.It is observed that the proposed schemes and the precoded system are both able to extract the full diversity and achieve the same diversity order.However,OBPL and OPL systems are able to outperform the precoded system by 2.6dB and 1.5dB,respectively,at BER=10?4.On the other hand,OBL remains inferior to the precoded system by ～1dB.It should be further emphasized that,besides performance improvements,OBPL and OPL have advantage over the precoded systems in terms of receiver complexity.The receiver complexity of proposed algorithms is independent of the channel length while the detector complexity in precoded systems is exponentially proportional to the channel length [28].

Another comparison in Fig.6is with the power loading scheme proposed in [20]by Hajiaghayi et.al.which also aims to optimize the BER performance (named as HDL scheme in our ?gure).Our results illustrate the superiority of proposed algorithms over HDL scheme.Speci ?cally at BER=10?3,we observe that OBL,OPL and OBPL outperform HDL by 0.8dB,3.2dB and 4.3dB respectively.It should be noted that HDL scheme is derived under the high SNR assumption and considers identical subcarrier power at source and relay nodes.Our schemes avoid such restricting assumptions and are therefore able to provide a better performance.Example 3(Effect of Relay Location)

In this example,we study the effect of relay location on the performance of EBPL,OBL,OPL,and OBPL schemes.Under the assumption of average transmission rate of 2bits/subcarrier and a target ?xed BER of 10?3,we plot the required SNR versus the relay location.Fig.7illustrates that the best location (i.e.requires the lowest SNR to achieve the BER =10?3)for OBPL and OPL is near to destination.For OBL and EBPL,mid-locations between source and destination become more favourable.For near-to-destination and near-to-source locations,they exhibit identical channel statistical properties [29]and therefore yield symmetric performance around 0dB location.

Fig.7.Effect of relay location on different adaptive algorithms. Example4(Effect of Channel Estimation)

In this example,we study the effect of channel estimation

on the proposed schemes.The adaptive algorithms assume that perfect CSI is available at source,relay and destination

nodes.In practice,CSI information needs to be estimated.

CSI for the direct link(i.e.,S→D)and relaying link(i.e., S→R and R→D)is also used at the destination for detection

process.For the estimation of relaying path,we adopt the so-called disintegrated channel estimation(D-CE)approach[29]

in which S→R and R→D channels are estimated separately.

In this approach,the relay node is equipped with a channel estimator and feed-forwards the S→R channel estimate to the

destination terminal as well as feedbacks it to the source.

Channel estimates for S→D and R→D links are obtained at the destination which sends them to the source and the relay

via a feedback channel.

In Fig.8,we assume the employment of linear minimum mean squared error estimator(LMMSE)[30]and perfect

feedback of the estimates.We consider the case where the relay is near to the destination,i.e.,G SR/G RD=?30dB and an average transmission rate of2bits/subcarrier.Note

that the precoded system which is used as a benchmark does not need CSI at the transmitter side.Fig.8illustrates that,at BER=10?4,OBPL and OPL schemes with imperfect channel estimation at the transmitter side are still able to outperform the precoded system by2.3dB and0.9dB.

In Fig.9,we study the effect of?nite-rate feedback which

is required to transfer the quantized CSIs in the practical implementation of our proposed schemes.Fig.9shows that we need as small as5bits to achieve a similar performance to the ideal system with perfect feedback.When6bits is used, it gives an identical performance to that of perfect feedback. The corresponding performance curve is not included in the ?gure for the sake of presentation.

V.C ONCLUSION

In this paper,we have investigated adaptive bit and/or power loading for a cooperative OFDM system with AF relaying.We have adopted BER as the objective function and

Fig.8.Performance of OBPL,OPL,OBL algorithms with imperfect channel estimation.

Fig.9.Effect of?nite-rate feedback on the performance of OPBL algorithm.

formulated three optimization problems leading to different adaptive algorithms.The?rst algorithm,named as OPL,com-putes the optimal source and relay power loading coef?cients under total power constraint and?xed subcarriers rate.The second algorithm,named as OBL,computes the optimal bit loading coef?cients under?xed average transmission rate and equal power loading.The third one,named as OBPL, computes the joint optimal power and bit loading coef?cients. Through Monte-Carlo simulations,we have demonstrated the superiority of our schemes over conventional non-adaptive cooperative OFDM systems.For example,assuming relay is located close to the destination,OBPL outperforms non-adaptive system(i.e.,equal bit and power loading)by5.8dB at a target BER=10?3.OBPL and OPL systems are also able to outperform the precoded cooperative OFDM systems while OBL turns out to be inferior.This also indicates that power loading is more dominant in the performance optimization where continuous optimization gives more degrees of freedom than the discrete optimization deployed in bit loading.We have further provided simulation results to quantify the effect of relay location,imperfect channel estimation and?nite-

rate quantized feedback on the BER performance of proposed schemes.

A PPENDIX

In this appendix,we provide the convexity proofs for OPL,OBL and OPBL optimization problems under consideration.OPL

For this problem,the general objective function given by (8)reduces to

P ≈c

N ∑k =1

Q (√

aεk ).

(41)

Let f 1=Q (√aεk )

.Since a and εk are positive quantities,f 1is a convex function.We ?rst need to prove that εk is concave in E k,S and E k,R .This will prove the convexity of f 1with respect to E k,S and E k,S [31].Rewriting εk given by (9)as

εk =g 1(E k,S ,E k,R )+g 2(E k,S ,E k,R )(42)where g 1(E k,S ,E k,R )=E k,S ∣D k,1∣2and

g 2(E k,S ,E k,R )=? ? ?E k,S E k,R ∣D k,2∣2(E k,R D k,R +D k,0)

orig .E k,S E k,R ∣D k,2∣2

(E k,R D k,R +E k,S G SR ∣D SR (k,k )∣2

+N 0)mod .(43)

In (43),we consider two cases.As earlier discussed,the

original IPS scaling term is given by (3).The modi ?ed version involves replacing E k,S with the average value E in (3).

Modi ?ed IPS:For the modi ?ed versiom,g 2(E k,S ,E k,R )

can be expressed as g 2(E k,S ,E k,R )=(∣D k,2∣2/D k,R D k,0

)(1E k,R E k,S D k,R +1D k,0E k,S

)(44)Since D k,2,D k,0,D k,R and E k,S are positive quantities g 2(E k,S ,E k,R )is concave because 1/(E k,R E k,S D k,R )and 1/(D k,0E k,S )are convex functions.The Hessian matrix of

H 1is expressed as

H 1=1D k,R E k,S E k,R [2E 2k,R

1E k,S E k,R 1E k,S E k,R 2

E 2k,S

](45)Since D k,R is positive,therefore H 1is positive de ?nite matrix which proves the convexity of [1/(E k,R E k,S D k,R )]

term.Thus g 2(E k,S ,E k,R )is a concave function.Original IPS:For the original case,(43)can be expressed as

g 2(E k,S ,E k,R )=4∣D k,2∣

N 20D k,R G SR ∣D SR (k,k )∣

21E k,R D k,R +N 02+1

E k,S G SR ∣D SR (k,k )∣2

+N 02(46)

The Hessian matrix H 2of the term in denominator is given

H 2=??2G SR ∣D SR (k,k )∣2

(E k,S G SR ∣D SR (k,k )∣2

+N 02)3002D k,R

(E k,R D k,R +N 02)

3??(47)Since G SR ,D k,R ,N 0and ∣D SR (k,k )∣are positive val-ues,H 2is a positive de ?nite matrix which proves the con-vexity term of the denominator term in (46).Accordingly

g 2(E k,S ,E k,R )is a concave function.As a result scaling with E k,S or E does not affect the concavity nature of g 2(E k,S ,E k,R ).Since g 1(E k,S ,E k,R )and g 2(E k,S ,E k,R )are concave functions,

εk is also concave function.As a result,f 1=Q (√aεk )is convex.Finally,noting that the sum of convex functions is convex,(41)turns out to be convex.OBL

For this problem,the objective function will follow the general form in (8).By using the values of a k and c k for M-PSK in (10)and (11),we have

P ≈N ∑k =1

(2b k ?1)mN Q (√2N 0sin 2(π2b k

)εk )(48)

By relaxing the integer variable b k to be continuous [26]and de ?ning

f 2=

(2b k ?1)mN Q (√2N 0sin 2

(π2b k )εk )(49)then P is convex if f 2is convex.To test the convexity of f 2,we obtain the second derivative with respect to b k (which was omitted here due to space limitations)which can be shown to be positive.Then as a result,f 2is convex and accordingly P

is convex.

OBPL Replacing a k and c k in for M-PSK in the general form of

the objective function in,we have

P =N ∑k =1

(2b k ?1)mN Q (√2N 0sin 2

(π2b k

)εk (E k,S ,E k,R ))(50)De ?ne f 3as

f 3=(2b k ?1)mN Q (√2

N 0sin 2(π2b k

)εk (E k,S ,E k,R )

)(51)

To prove the convexity of P ,we need to prove that f 3

is convex.First note that sin (π/2b k )can be bounded as 0≤sin (π/2b k )≤1.Noting Q function ′s monotonically decreasing property,it is straightforward to show that

(2b k ?1)mN Q (√2

N 0εk (E k,S ,E k,R ))≤f 3≤(2b k ?1)2mN

(52)Further de ?ne f 4

=(2b k ?1)/

(2mN )and f 5=((2b k ?1)/mN )Q (√(2/N 0)εk (E k,S ,E k,R )),f 5is con-vex function if f 3and f 4are convex functions.We know that f 4is convex function,thus we need to show that f 5is convex.f 5can be bounded as f 6

(2/N 0)εk (E k,S ,E k,R )

)and f 7=(1/2mN )exp (b k ?(1/N 0)εk (E k,S ,E k,R )).Noting exp (.)is a convex non-decreasing function and (b k ?(1/N 0)εk (E k,S ,E k,R ))is convex,is convex in (b k ,E k,S ,E k,R )[31].Since f 6and f 7are convex,f 5is convex and accordingly f 3is convex given that P of the OBPL is convex in (b k ,E k,S ,E k,R ).

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2004.

Osama Amin received B.Sc.degree in Electrical

and Electronics Engineering from South Valley Uni-

versity,Aswan,Egypt,in2000and M.Sc.degree

in Electrical Engineering from Assiut University,

Assiut,Egypt in2004.He is currently a Ph.D.

student in Electrical and Computer Engineering,

University of Waterloo,Canada.His research inter-

ests include cooperative communications,adaptive

systems,OFDM,and channel

estimation.

Murat Uysal was born in Istanbul,Turkey in1973.

He received the B.Sc.and the M.Sc.degree in

electronics and communication engineering from

Istanbul Technical University,Istanbul,Turkey,in

1995and1998,respectively,and the Ph.D.degree in

electrical engineering from Texas A&M University,

College Station,Texas,in2001.Since2002,he has

been with the Department of Electrical and Com-

puter Engineering,University of Waterloo,Canada,

where he is now an Associate Professor.He is

currently on leave at¨Ozye?g in University,Istanbul, Turkey.His general research interests lie in communications theory and signal processing for communications with special emphasis on wireless appli-cations.Speci?c research areas include MIMO communication techniques, space-time coding,diversity techniques and coding for fading channels, cooperative communication,and free-space optical communication.

Dr.Uysal is an Associate Editor for IEEE T RANSACTIONS ON W IRELESS C OMMUNICATIONS and IEEE C OMMUNICATIONS L ETTERS.He was a Guest Co-Editor for the Wiley Journal on Wireless Communications and Mobile Computing Special Issue on“MIMO Communications”(October 2004),and IEEE J OURNAL ON S ELECTED A REAS IN C OMMUNICATIONS Special Issue on“Optical Wireless Communications”(December2009).Over the years,he has served on the technical program committee of more than 60international conferences in the communications area.He recently co-chaired IEEE ICC’07Communication Theory Symposium and CCECE’08 Communications and Networking Symposium.Dr.Uysal is a Senior IEEE member.

复习思考题完整版(2019.04)

复习思考题第1章绪论 1.机械原理课程的研究对象是什么？其研究内容有哪几方面？ 2.机器和机构有何联系与区别？构件和零件有何联系与区别？ 3.试列举3个机构的实例，并说明其组成与功能。 4.试列举3个机器的实例，并说明其组成与功能。第2章机构的结构分析 5.构件自由度的定义？机构自由度的定义？ 6.运动副及运动副元素的定义？运动副的分类？平面高副与平面低副的区别？ 7.运动链的定义？机构具有确定运动（运动链成为机构）的条件？ 8.构件数、约束数与机构自由度的关系（公式）？ 9.什么是局部自由度？复合铰链？虚约束？应如何处置？ 10.什么是基本杆组？基本杆组的自由度是多少？基本杆组中运动构件数与低副是如何匹配的？ 11.机构的组成原理是什么？机构的结构分析步骤如何？ 12.为什么要进行高副低代？高副低代的条件？一个高副需要用什么来替代？第3章平面连杆机构及其设计 13.等腰梯形机构、平行四边形及反平行四边形机构各属于什么机构？有什么特征？ 14.平面四杆机构的演化有哪几种方式？试分别举例说明平面四杆机构的演化？ 15.铰链四杆机构曲柄存在（转动副成为整转副）的条件是什么？ 16.曲柄摇杆机构中，当以曲柄为原动件时，机构是否一定存在急回运动，且一定无死点？为什么？ 17.曲柄滑块机构在什么情况下会出现急回特性？ 18.四杆机构中的极位和死点有何异同？ 19.行程速比系数K的大小取决于什么（公式）？与急回特性之间的关系怎样？

20.压力角的定义？与传动角的关系？四杆机构的最小传动角可能出现在什么地方？第4章凸轮机构及其设计 21.凸轮机构的有几种分类方式？各有什么特征？ 22.描述凸轮机构工作情况的参数有哪些？ 23.何谓凸轮机构传动中的刚性冲击和柔性冲击？常用的从动件运动规律中哪些产生刚性冲击？哪些产生柔性冲击？ 24.为什么要进行从动件运动规律的组合？应满足什么条件？ 25.反转法设计凸轮廓线的设计思想？ 26.何谓凸轮机构的压力角？它在哪一轮廓曲线上度量？压力角变化对凸轮机构的工作有何影响？与凸轮几何尺寸有何关系？ 27.滚子半径的选择与理论轮廓曲率半径有何关系？如实际轮廓曲线出现失真，应该采取哪些措施？ 28.从动件的偏置方向应根据什么原则？为什么？第5章齿轮机构及其设计 29.为了实现定传动比传动，对齿轮的齿廓曲线有什么要求？ 30.为什么说啮合线是“四线合一”？“四线合一”对齿轮传动具有什么好处？ 31.渐开线标准直齿圆柱齿轮的分度圆具有哪些特征？渐开线标准直齿圆柱齿轮标准安装时具有哪些特点？ 32.什么是渐开线齿廓的运动可分性？为什么它对传动有利？ 33.渐开线标准齿轮的基本参数是指哪些？主要几何尺寸如何计算（公式）？ 34.渐开线直齿轮、斜齿轮和圆锥齿轮的正确啮合条件、连续传动条件各是什么？ 35.何谓重合度？重合度的大小与哪些因素有关（公式）？ 36.分度圆和节圆、啮合角与压力角有何区别？分度圆和节圆在什么情况下相重合？ 37.何谓根切？有何危害？造成根切的原因是什么？ 38.齿轮为什么要进行变位？正变位齿轮、标准齿轮和负变位齿轮的在齿形

材料科学基础习题与答案

第二章思考题与例题 1. 离子键、共价键、分子键和金属键的特点，并解释金属键结合的固体材料的密度比离子键或共价键固体高的原因 2. 从结构、性能等方面描述晶体与非晶体的区别。 3. 何谓理想晶体何谓单晶、多晶、晶粒及亚晶为什么单晶体成各向异性而多晶体一般情况下不显示各向异性何谓空间点阵、晶体结构及晶胞晶胞有哪些重要的特征参数 4. 比较三种典型晶体结构的特征。（Al、α-Fe、Mg三种材料属何种晶体结构描述它们的晶体结构特征并比较它们塑性的好坏并解释。）何谓配位数何谓致密度金属中常见的三种晶体结构从原子排列紧密程度等方面比较有何异同 5. 固溶体和中间相的类型、特点和性能。何谓间隙固溶体它与间隙相、间隙化合物之间有何区别（以金属为基的）固溶体与中间相的主要差异（如结构、键性、性能）是什么 6. 已知Cu的原子直径为A，求Cu的晶格常数，并计算1mm3Cu的原子数。 7. 已知Al相对原子质量Ar（Al）=，原子半径γ=，求Al晶体的密度。 8 bcc铁的单位晶胞体积，在912℃时是；fcc铁在相同温度时其单位晶胞体积是。当铁由bcc转变为fcc时，其密度改变的百分比为多少 9. 何谓金属化合物常见金属化合物有几类影响它们形成和结构的主要因素是什么其性能如何

10. 在面心立方晶胞中画出[012]和[123]晶向。在面心立方晶胞中画出（012）和（123）晶面。 11. 设晶面（152）和（034）属六方晶系的正交坐标表述，试给出其四轴坐标的表示。反之，求（3121）及（2112）的正交坐标的表示。（练习），上题中均改为相应晶向指数，求相互转换后结果。 12.在一个立方晶胞中确定6个表面面心位置的坐标，6个面心构成一个正八面体，指出这个八面体各个表面的晶面指数，各个棱边和对角线的晶向指数。 13. 写出立方晶系的{110}、{100}、{111}、{112}晶面族包括的等价晶面，请分别画出。 14. 在立方晶系中的一个晶胞内画出（111）和（112）晶面，并写出两晶面交线的晶向指数。 15 在六方晶系晶胞中画出[1120]，[1101]晶向和（1012）晶面，并确定（1012）晶面与六方晶胞交线的晶向指数。 16.在立方晶系的一个晶胞内同时画出位于（101），（011）和（112）晶面上的[111]晶向。 17. 在1000℃，有W C为%的碳溶于fcc铁的固溶体，求100个单位晶胞中有多少个碳原子（已知：Ar(Fe)=，Ar(C)=） 18. r-Fe在略高于912℃时点阵常数a=，α-Fe在略低于912℃时a=，求：（1）上述温度时γ-Fe和α-Fe的原子半径R；（2）γ-Fe→α-Fe转变时的体积变化率；（3）设γ-Fe→α-Fe转变时原子半径不发生变化，求此转变时的体积变

英语花费四种用法的区别

spend的主语必须是人，常用于以下结构： (1)sb. spend time ／money on sth. 在……上花费时间(金钱)。 (2) sb.spend time ／ money (in) doing sth.花费时间(金钱)做某事。例：I spent fifty yuan on the coat. = I spent fifty yuan (in) buying the coat. 我花50元买了这件大衣。 He spent three days on the work. = He spend three days (in) doing the work. 我干这项工作用了3天。 (3). spend money for sth. 花钱买例如： His money was spent for books. 他的钱用来买书了。 cost的主语是物或某种活动，还可以表示"值"，常见用法如下： (1)sth. costs (sb.) ＋金钱,某物花了(某人)多少钱。例：A new computer costs a lot of money. 买一台新电脑要花一大笔钱。 (2) (doing) sth. costs (sb.) ＋时间,某物(做某事)花了(某人)多少时间。例：Remembering these new words cost him a lot of time. 他花了大量时间才记住了这些单词。注意：cost的过去式及过去分词都是cost,并且不能用于被动句。 take后面常跟双宾语，常见用法有以下几种： (1) It takes sb. ＋时间＋to do sth. 做某事花了某人多少时间。例：It took them three years to build this road. 他们用了三年时间修完了这条路。 (2)doing sth. takes sb. ＋时间,做某事花了某人多少时间。例：Repairing this car took him the whole afternoon. 他花了一下午修车。 pay的基本用法是： (1) pay (sb.) money for sth. 付钱(给某人)买……。例：I have to pay them 20 pounds for this room each month. 我每个月要付20英磅的房租。 (2)pay for sth. 付……的钱。例：I have to pay for the book lost. 我不得不赔丢失的书款。 (3)pay for sb.替某人付钱。例：Don‘t worry!I'll pay for you. 别担心，我会给你付钱的。 (4)pay sb.付钱给某人。例: They pay us every month.他们每月给我们报酬。 (5)pay money back 还钱。(6)pay off one's money还清钱。例：May I borrow 12 yuan from you? I'll pay it back next week. 你能借给我12块钱吗？下周还你。

(完整版)复习思考题

复习思考题 1. 何谓内环境？内环境为什么要保持相对稳定？ 2. 生理功能调节的方式有哪些？并比较其异同。 3. 体内的控制系统有哪几类？并比较其异同。 4. 何谓反馈、正反馈、负反馈、前馈？ 5. 反应、反射和反馈有何区别？ 6. 生理学的研究方法有哪些？可从哪些水平研究？ 1.简述细胞膜物质转运有哪些方式？ 2.Na+-K+泵的作用意义？ 3.在一般生理情况下,每分解一分子ATP,钠泵运转可使 1. 细胞间通讯有哪些方式？各种方式之间有何不同？ 2. 通过细胞表面受体介导的跨膜信号转导有哪几种方式？比较各种方式之间的异同。 3. 试述细胞信号转导的基本特征。 4. 试比较G蛋白偶联受体介导的几种信号通路之间的异同。 5. 概述受体酪氨酸介酶介导的信号通路的组成、特点及其主要功能。

1.静息电位产生的原理是什么？如何证明？ 2.动作电位是怎么发生的？如何证明动作电位是钠的平衡电位？ 3.发生兴奋过程中，如何证明有兴奋性的变化？为什么会发生这些变化？ 4.兴奋是如何传导的？影响传导速度的因素有哪些？ 5.试比较局部电位和动作电位的区别。 6.刺激引起神经兴奋的内因和外因是什么？ 7.绝对不应期是否指潜伏期？潜伏期是否等于引起兴奋所需的最短刺激作用时间？ 8.神经干上某点发生兴奋后，除向前传导外能否逆传？为什么？ 9.试比较改变刺激强度，单一神经纤维与神经干的动作电位变化？为什么？ 10.血K+浓度对兴奋性、RP和AP有何影响？ 11、以下关于细胞膜离子通道的叙述,正确的是(c ) A.在静息状态下,Na、K离子通道都处于关闭状态 B.细胞受刺激刚开始去极化时,钠离子通道就大量开放 C.在动作电位去极相,钾离子通道也被激活,但出现较慢 D.钠离子通道关闭,出现动作电位的复极相 E.钠、钾离子通道被称为化学依从性通道

现在完成时的用法解析

Lead in 一般过去时导入：一般过去时的概念一般过去时表示过去某个时间发生的动作或存在的状态。 I lost my pen.(过去时) I have lost my pen.（现在完成时）过去时一般表示过去所发生的动作或存在的状态。“我把笔丢了”，说话人只是想告诉对方在过去的某一个时间里他的笔丢了，仅此而已。（仅仅是在陈诉过去的一个事实，与现在没有关系） New lesson 一．现在完成时的构成：助动词have/has(not) +动词过去分词 She has turned on the lights. 肯定式：have/ has+过去分词否定式：haven’t/hasn’t+过去分词疑问式：把have/has 提到主语前面肯定回答：Yes,…have/has 否定回答：No,…..haven’t/hasn’t 二．用法 1.表示过去发生的动作对现在造成的影响或结果. 例如:She has gone. 她走了. She went in the past.(She is not here

now.) I have just cleaned my hands. 我刚洗过手。（“洗手”是发生在过去的动作，对现在造成的结果是“手干净了”） I have closed the door. (The door is close now.) She has turned on the lights. (The lights are on now.) He has written his name on the blackboard. (The name is on the blackboard now.) Mother has cooked the dinner. (We can have dinner now.) You have seen the film. （You know what the film is about.) We have cleaned the classroom. (The classroom is clean now.) 2. 表示动作发生在过去，并且一直延续到现在，甚至还可能继续延续下去。这时常和since所引导的短语或从句或for引导的短语连用(for有时可以省去)。以及so far, by now, these days, in the last/past … years/days…等连用。 I have lived here for over ten years. （表示他现在还在这里居住，并且还有可能继续住下去）比较：I lived here ten years ago.(仅仅说明了他十年前在这里记住过的事实,不代表现在他还住在这里) He has studied here since 2006. 都用How long 提问，例： We have learned English for three years. We have learned English since three years ago.

《材料科学与工程基础》习题和思考题及答案

《材料科学与工程基础》习题和思考题及答案第二章 2-1.按照能级写出N、O、Si、Fe、Cu、Br原子的电子排布（用方框图表示）。 2-2.的镁原子有13个中子，11.17%的镁原子有14个中子，试计算镁原子的原子量。 2-3.试计算N壳层内的最大电子数。若K、L、M、N壳层中所有能级都被电子填满时，该原子的原子序数是多少？ 2-4.计算O壳层内的最大电子数。并定出K、L、M、N、O壳层中所有能级都被电子填满时该原子的原子序数。 2-5.将离子键、共价键和金属键按有方向性和无方向性分类，简单说明理由。2-6.按照杂化轨道理论，说明下列的键合形式：（1）CO的分子键合（2）甲烷CH的分子键合 24 （3）乙烯CH的分子键合（4）水HO的分子键合 242 （5）苯环的分子键合（6）羰基中C、O间的原子键合 2-7.影响离子化合物和共价化合物配位数的因素有那些？ 2-8.试解释表2-3-1中，原子键型与物性的关系？ 332-9.0?时，水和冰的密度分别是1.0005 g/cm和0.95g/cm，如何解释这一现象？ +2-10.当CN=6时，K离子的半径为0.133nm(a)当CN=4时，半径是多少？(b)CN=8时，半径是多少？

32-11.(a)利用附录的资料算出一个金原子的质量？(b)每mm 的金有多少个原子？(c)根据金 21的密度，某颗含有10个原子的金粒，体积是多少？(d)假设金原子是球形 (r=0.1441nm),Au21并忽略金原子之间的空隙，则10个原子占多少体积？(e)这些金原子体积占总体积的多少百分比？ 2+2-2-12.一个CaO的立方体晶胞含有4个Ca离子和4个O离子，每边的边长是0.478nm，则CaO的密度是多少？ 2-13.硬球模式广泛的适用于金属原子和离子，但是为何不适用于分子？ 2-14.计算（a）面心立方金属的原子致密度；（ b）面心立方化合物NaCl的离子致密度（离子半径r+=0.097，r-=0.181）；（C）由计算结果，可以引出什么结论？ NaCl 470 2-15.铁的单位晶胞为立方体，晶格常数 a=0.287nm，请由铁的密度算出每个单位晶胞所含的原子个数。 2-16.钛的单位晶胞含有两个原子，请问此单位晶胞的体积是多少？ 2-17.计算面心立方、体心立方和密排六方晶胞的致密度。 2-18.在体心立方结构晶胞的（100）面上按比例画出该面上的原子以及八面体和四面体间隙。 2-19.键合类型是怎样影响局部原子堆垛的？ 2-20.厚度0.08mm、面积670mm2的薄铝片(a)其单位晶胞为立方体， a=0.4049nm，则此薄片

四种花费和四种提供的用法

英语中“花费”的四种用法王朝红的工作室英语花费四种用法 spend的主语通常是人，往往用于以下句型： 1. (sb) spend some money/some time on sth。 2. (sb)spend some money/some time(in)doing sth。例如： I spent fifty yuan on the coat。 = I spent fifty yuan (in) buying the coat. 我花50元买了这件大衣。 He spent three days on the work. = He spend three days (in) doing the work. 我干这项工作用了3天。 3.spend money for sth. 花钱买……。例如：His money was spent for books. 他的钱用来买书了。 take常用于“占用、花费”时间，后面常跟双宾语，其主语通常为形式主语“it”或物。句式是： 1. It takes/took sb.some time to do sth 例如：It will take me two days to do the work. 这项工作花了2天时间。 2. Doing sth./Sth.takes sb.some time. 例如： The work will take me two days。这项工作花了2天时间。

Repairing this car took him the whole afternoon. 他花了一下午修车。 It took me three years to draw the beautiful horses。 =Drawing the beautiful horses took me three years。画这些漂亮的马花费了我3年时间。 pay为“付款、赔偿”之意，主语通常是人，句型 1. sb. pays some money for sth 例如： I paid fifty yuan for the coat。我花50元买了这件大衣。 2. pay (sb。) money for sth. 付钱(给某人)买……。例如：I have to pay them 20 pounds for this room each month. 我每个月要付20英磅的房租。 3. pay money back 还钱。例如：May I borrow 12 yuan from you? I'll pay it back next week. 你能借给我12块钱吗?下周还你。 4. pay off one's money 还清钱。 cost的主语是物或某种活动，还可以表示“值”。句型 1. sth. costs (sb。) +money, 某物花了(某人)多少钱。

完整版机器视觉思考题及其答案

什么是机器视觉技术？试论述其基本概念和目的。答：机器视觉技术是是一门涉及人工智能、神经生物学、心理物理学、计算机科学、图像处理、模式识别等诸多领域的交叉学科。机器视觉主要用计算机来模拟人的视觉功能，从客观事物的图像中提取信息，进行处理并加以理解，最终用于实际检测、测量和控制。机器视觉技术最大的特点是速度快、信息量大、功能多。机器视觉是用机器代替人眼来完成观测和判断，常用于大批量生产过程汇总的产品质量检测，不适合人的危险环境和人眼视觉难以满足的场合。机器视觉可以大大提高检测精度和速度，从而提高生产效率，并且可以避免人眼视觉检测所带来的偏差和误差。机器视觉系统一般由哪几部分组成？试详细论述之。答：机器视觉系统主要包括三大部分：图像获取、图像处理和识别、输出显示或控制。图像获取：是将被检测物体的可视化图像和内在特征转换成能被计算机处理的一系列数据。该部分主要包括，照明系统、图像聚焦光学系统、图像敏感元件（主要是CCD和CMOS采集物体影像。图像处理和识别：视觉信息的处理主要包括滤波去噪、图像增强、平滑、边缘锐化、分割、图像识别与理解等内容。经过图像处理后，图像的质量得到提高，既改善了图像的视觉效果又便于计算机对图像进行分析、处理和识别。输出显示或控制：主要是将分析结果输出到显示器或控制机构等输出设备。试论述机器视觉技术的现状和发展前景。答：。机器视觉技术的现状：机器视觉是近20?30年出现的新技术，由于其固有的柔性好、非接触、快速等特点，在各个领域得到很广泛的应用，如航空航天、工业、军事、民用等等领域。发展前景：随着光学传感器、信息技术、信号处理、人工智能、模式识别研究的不断深入和计算机性价比的不断提高，机器视觉技术越来越成熟，特别是市面上已经有针对机器视觉系统开发的企业提供配套的软硬件服务，相信越来越多的客户会选择机器视觉系统代替人力进行工作，既便于管理又节省了成本。价格持续下降、功能逐渐增多、成品小型化、集成产品增多。机器视觉技术在很多领域已得到广泛的应用。请给出机器视觉技术应用的三个实例并叙述之。答：一、在激光焊接中的应用。通过机器视觉系统，实时跟踪焊缝位置，实现实时控制，防止偏离焊缝，造成产品报废。二、在火车轮对检测中的应用，通过机器视觉系统抓拍轮对图像，找出轮对中有缺陷的轮对，提高检测精度和速度，提高效率。三、大批量生产过程中的质量检查，通过机器视觉系统，对生产过程中的产品进行质量检查跟踪，提高生产效率和准确度。什么是傅里叶变换，分别绘出一维和二维的连续及离散傅里叶变换的数学表达式。论述图像傅立叶变换的基本概念、作用和目的。答：傅里叶变换是将时域信号分解为不同频率的正弦信号或余弦函数叠加之和。一维连续函数的傅里叶变换为：一维离散傅里叶变换为：二维连续函数的傅里叶变换为：二维离散傅里叶变换为：图像傅立叶变换的基本概念：傅立叶变换是数字图像处理技术的基础，其通过在时空域和频率域来回切换图像，对图像的信息特征进行提取和分析，简化了计算工作量，被喻为描述图像信息的第二种语言，广泛应用于图像变换，图像编码与压缩，图像分割，图像重建等。作用和目的：图像的频率是表征图像中灰度变化剧烈程度的指标，是灰度在平面空间上的梯度。傅立叶变换的物理意义是将图像的灰度分布函数变换为图像的频率分布函数，傅立叶逆变换是将图像的频率分布函数变换为灰度分布函数。图像灰度变换主要有哪几种形式？各自的特点和作用是什么？答：灰度变换:基于点操作，将每一个像素的灰度值按照一定的数学变换公式转换为一个新的灰度值。灰度变换是图像增强的一种重要手段，它可以使图像动态范围加大，使图像的对比度扩展，

(英语语法)四种完成时态

LESSON EIGHT 四种完成时态主系表现在：You are rich. 过去：You were rich. 将来：You will be rich. 过去将来：You would be rich. There be 现在：There is a book on the desk . 过去：There was a book on the desk. 将来：There will be a book on the desk. 过去将来：There would be a book on the desk.主谓宾状现在：You study English in the school. 过去时：You studied English in the school. 将来时: You will study English in the school. You are going to study English. You are to study English. are about to study English would study English in the scho You were going to study English. You were to study English. You were about to study English. You are studying English. You were studying English. You will be studying English. You woud be studying English.

环境材料学课后思考题教程文件

环境材料学课后思考题

1 用自己的理解给出生态环境材料的定义。答：生态环境材料是指那些具有满意的使用性能和可接受的经济性能，并在其制备、使用及废弃过程中对资源和能源消耗较少，对生态环境影响较小且再生利用率较高的一类材料。 1 你认为那些材料属于生态环境材料？举例说明。（举例之后还要简要说明一下）答：比如：生态水泥、环保建材、降解树脂、环境工程材料天然资源环境材料、电磁波防护类材料、电子功能材料领域的毒害元素替代材料 2 试用物质不灭和能量守恒的理论来说明材料与资源、环境的关系答：众所周知，材料的生产往往要消耗大量的资源。当生产效率一定时，除有效产品外，大量的废弃物被排放到环境中去，造成了环境的污染。因此对材料的生产和使用而言，资源消耗是源头，环境污染是末尾。也就是说，材料的生产和使用与资源和环境有密不可分的关系。 2 图2-1是一个典型的开环工业生产链，从环保的角度看，若能实现闭环的工业生产链，可明显减少废弃物排放。请选择一个你感兴趣的产品设计一个闭环生产流程。资能源资能源资能源

铁矿铁水钢胚成品（热轧钢板等）使用废弃污染物污染物污染物 4 选择一个你所熟悉的材料产品或过程，用物质流方法进行资源效率分析，并就如何提高资源效率提出具体的技术措施。答：钢铁的资源效率高达10.4%，就是生产1t纯金属材料所消耗的原材料将近12t。具体技术措施：1）由外界向高炉-转炉流程内某中间工序输入废钢，可提高流程铁资源效率；2）由外界向高炉-转炉流程内某中间工序输入铁矿石等自然铁资源，可提高该流程铁资源效率。3）通过提高电炉钢比，提高铁资源效率。 5 根据你对可持续发展的理解，考虑如何实现（1）金属材料，（2）高分子材料，（3）无机非金属材料（选一种）的可持续发展，并提出几项可具体实施的技术措施。金属材料：利用微生物冶金；代替含稀缺合金元素的新型合金材料和不含毒害元素的材料，以及废弃物无害资源化转化技术；少合金化与通用合金，形成绿色/生态材料体系，有利于材料的回收与再生利用。高分子材料：1）建立必要的法规，加强全民的环保意识；2）回收塑料，变废为宝：燃烧废旧塑料利用热能，热分解提取化工原料和改进操作技术、设备3）发展环境友好高分子：可降解塑料的开发和合成，采用生物发酵的方法合成的生物高分子。

“四个花费”spend,cost,take, pay讲解及对应中考练习

“四个花费”讲解及对应中考练习 spend，cost，take, pay spend，cost，take和pay的区别是历年考试的必考内容之一，虽然它们都可以表示"花费"，但用法却不尽相同，讲解如下： spend的主语必须是人，常用于以下结构： (1) spend time ／money on sth. 在……上花费时间(金钱)。例：I spent two hours on this maths problem. 这道数学题花了我两个小时。 (2) spend time ／money (in) doing sth. 花费时间(金钱)做某事。例：They spent two years (in) building this bridge. 造这座桥花了他们两年时间。 (3)spend money for sth. 花钱买……。例：His money was spent for books. 他的钱用来买书了。 cost的主语是物或某种活动，还可以表示"值"，常见用法如下： (1)sth. costs (sb.) ＋金钱,某物花了(某人)多少钱。例：A new computer costs a lot of money. 买一台新电脑要花一大笔钱。 (2) (doing) sth. costs (sb.) ＋时间,某物(做某事)花了(某人)多少时间。例：Remembering these new words cost him a lot of time. 他花了大量时间才记住了这些单词。注意：cost的过去式及过去分词都是cost,并且不能用于被动句。 take后面常跟双宾语，常见用法有以下几种： (1) It takes sb. ＋时间＋to do sth. 做某事花了某人多少时间。例：It took them three years to build this road. 他们用了三年时间修完了这条路。 (2)doing sth. takes sb. ＋时间,做某事花了某人多少时间。例：Repairing this car took him the whole afternoon. 他花了一下午修车。 pay的基本用法是： (1) pay (sb.) money for sth. 付钱(给某人)买……。例：I have to pay them 20 pounds for this room each month. 我每个月要付20英磅的房租。 (2)pay for sth. 付……的钱。例：I have to pay for the book lost. 我不得不赔丢失的书款。 (3)pay for sb. 替某人付钱。例：Don‘t worry!I'll pay for you. 别担心，我会给你付钱的。 (4)pay sb. 付钱给某人。例: They pay us every month.他们每月给我们报酬。 (5)pay money back 还钱。例：May I borrow 12 yuan from you? I'll pay it back next week. 你能借给我12块钱吗？下周还你。 (6)pay off one's money还清钱。【真题对应练习】 1.（广州某校2015期中）--- How much does your new bike ___________ ?

思考题整理完整版

教育资料

容器几何尺寸：（1）容器的大小；（2）形状h/D；h/D为0.25时杀菌时间最短。导热型圆罐的杀菌时间(扎丹)：t0=A(8.3hD+D2) 8.什么是致死率及部分杀菌量？致死率：致死率是热力致死时间的倒数，热力致死时间Ti的倒数1/Ti为在温度θi 杀菌1min所取得的效果占全部杀菌效果的比值,称为致死率. （以热处理时间为横坐标，以致死率为纵坐标图为致死率图。）部份杀菌量：细菌在T℃温度时的热力致死时间为I分钟，在T℃加热了t钟，则在T℃温度下完成的杀菌程度为t/τ。 9.说明比奇洛基本推算法的基本原理，并用图表示杀菌时间的推算方法。基本原理：找出罐头食品传热曲线和各温度时细菌热力致死时间性的关系，为罐头食品杀菌操作（理论上达到完全无菌程度）推算预定杀菌温度工艺条件下需要的加热冷却时间。（图自己补，分别是食品传热曲线，热力致死时间曲线，致死率曲线，三幅图加上文字表述） 9.杀菌方法的选择与酸度有什么关系？（网上找的）食品的酸度对微生物耐热性的影响很大。对绝大多数微生物来说，在pH中性范围内耐热性最强，pH升高或降低都可减弱微生物的耐热性。特别是在偏酸性时，促使微生物耐热性减弱作用更明显。酸度不同，对微生物耐热性的影响程度不同。同一微生物在同一杀菌温度，随着pH的下降，杀菌时间可以大大缩短。所以食品的酸度越高，pH越低，微生物及其芽胞的耐热性越弱。酸使微生物耐热性减弱的程度随酸的种类而异，一般认为乳酸对微生物的抑制作用最强，苹果酸次之，柠檬酸稍弱。由于食品的酸度对微生物及其芽胞的耐热性的影响十分显著，所以食品酸度与微生物耐热性这一关系在罐头杀菌的实际应用中具有相当重要的意义。酸度高，pH低的食品杀菌温度低一些，时间可短一些；酸度低，pH高的食品杀菌温度高一些，时间长一些。 10.为什么要进行反压冷却？如何进行操作？（网上找的）为减少冷却阶段罐内外压力差防止容器变形、损坏玻璃罐跳盖等现象，常采用反压冷却。加压冷却（反压冷却）：在通入冷却水的同时通入一定的压缩空气。（要注意的是，杀菌锅温度声高到了杀菌温度T，并不意味着罐内食品温度也达到了杀菌温度的要求，实际上食品尚处于加热升温阶段。对流传热型食品的温度在此阶段内常能迅速上升，甚至于到达杀菌温度。而导热型食品升温很慢，甚至于开始冷却时尚未能达到杀菌温度。因此冷却时需要加反压）操作：一般高温杀菌115～1210C，需打入137.3～166.7kPa的压力。杀菌釜内反压力的大小，以使杀菌釜内总压力（蒸汽压力与补充压力之和平等于或稍大于罐内压力与允许压力差Δp允的好，即： p釜＝p 釜蒸＋p反≥p2－Δp允 p反＝p2 －p釜蒸－Δp允反压杀菌冷却时所补充的压缩空气应使杀菌釜内压力恒定，一直维持到镀锡罐内压力降到1＋Δp允大气压，玻璃罐内压力降到常压时才可停止供给压缩空气。 11.说明内容物腐败变质的类型，分析其原因。胀罐：从程度分隐胀、轻胀、硬胀从性质分：理化性胀罐、细菌性胀罐。氢胀：［H+］↑→罐壁腐蚀→ H2↑假胀：装量过多，真空度低教育资料

材料力学思考题答案

材料力学复习思考题 1. 材料力学中涉及到的内力有哪些？通常用什么方法求解内力？轴力，剪力，弯矩，扭矩。用截面法求解内力 2. 什么叫构件的强度、刚度与稳定性？保证构件正常或安全工作的基本要求是什么？杆件的基本变形形式有哪些？构件抵抗破坏的能力称为强度。构件抵抗变形的能力称为刚度。构件保持原有平衡状态的能力称为稳定性。基本要求是：强度要求，刚度要求，稳定性要求。基本变形形式有：拉伸或压缩，剪切，扭转，弯曲。 3. 试说出材料力学的基本假设。连续性假设：物质密实地充满物体所在空间，毫无空隙。均匀性假设：物体内，各处的力学性质完全相同。各向同性假设：组成物体的材料沿各方向的力学性质完全相同。小变形假设：材料力学所研究的构件在载荷作用下的变形或位移,其大小远小于其原始尺寸。 4. 什么叫原始尺寸原理？什么叫小变形？在什么情况下可以使用原始尺寸原理？可按结构的变形前的几何形状与尺寸计算支反力与内力叫原始尺寸原理。可以认为是小到不至于影响内力分布的变形叫小变形。绝大多数工程构件的变形都极其微小，比构件本身尺寸要小得多，以至在分析构件所受外力（写出静力平衡方程）时可以使用原始尺寸原理。 5. 轴向拉伸或压缩有什么受力特点和变形特点。受力特点：外力的合力作用线与杆的轴线重合。变形特点：沿轴向伸长或缩短 6. 低碳钢在拉伸过程中表现为几个阶段？各有什么特点？画出低碳钢拉伸时的应力－应变曲线图，各对应什么应力极限。弹性阶段：试样的变形完全弹性的，此阶段内的直线段材料满足胡克定律εσE =。 p σ --比例极限。 e σ—弹性极限。屈服阶段：当应力超过b 点后，试样的荷载基本不变而变形却急剧增加，这种现象称为屈服。s σ--屈服极限。强化阶段：过屈服阶段后，材料又恢复了抵抗变形的能力，要使它继续变形必须增加拉力.这种现象称为材料的强化。b σ——强度极限局部变形阶段：过e 点后，试样在某一段内的横截面面积显箸地收缩，出现颈缩 (necking)现象，一直到试样被拉断。对应指标为伸长率和断面收缩率。 7. 什么叫塑性材料与脆性材料？衡量材料塑性的指标是什么？并会计算延伸率和断面收缩率。

四种花费和四种提供的用法

四种花费和四种提供的用法公司内部编号：（GOOD-TMMT-MMUT-UUPTY-UUYY-DTTI-

例如：It will take me two days to do the work. 这项工作花了2天时间。 2. Doing sth./Sth.takes sb.some time. 例如： The work will take me two days。这项工作花了2天时间。 Repairing this car took him the whole afternoon. 他花了一下午修车。 It took me three years to draw the beautiful horses。 =Drawing the beautiful horses took me three years。画这些漂亮的马花费了我3年时间。 pay为“付款、赔偿”之意，主语通常是人，句型 1. sb. pays some money for sth 例如： I paid fifty yuan for the coat。我花50元买了这件大衣。 2. pay (sb。) money for sth. 付钱(给某人)买……。例如：I have to pay them 20 pounds for this room each month.我每个月要付20英磅的房租。

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网络安全思考题答案第一章 1.什么是osi安全体系结构？为了有效评估某个机构的安全需求，并选择各种安全产品和策略，负责安全的管理员需要一些系统性的方法来定义安全需求以及满足这些安全需求的方法，这一套系统体系架构便称为安全体系架构。 2.被动和主动威胁之间有什么不同？被动威胁的本质是窃听或监视数据传输，主动威胁包含数据流的改写和错误数据流的添加。 3.列出并简要定义被动和主动安全攻击的分类？被动攻击：消息内容泄漏和流量分析。主动攻击：假冒，重放，改写消息和拒绝服务。 4.列出并简要定义安全服务的分类认证，访问控制，数据机密性，数据完整性，不可抵赖性。 5.列出并简要定义安全机制的分类。特定安全机制：为提供osi安全服务，可能并到适当的协议层中。普通安全机制：没有特定osi安全服务或者协议层的机制。第二章 1.对称密码的基本因素是什么？明文，加密算法，秘密密钥，密文，解密算法。

2.加密算法使用的两个基本功能是什么？替换和排列组合 3.分组密码和流密码区别是什么？流密码是一个比特一个比特的加密，分组时若干比特同时加密。比如DES是64bit的明文一次性加密成密文。密码分析方面有很多不同。比如说密码中，比特流的很多统计特性影响到算法的安全性。密码实现方面有很多不同，比如流密码通常是在特定硬件设备上实现。分组密码可以在硬件实现，也可以在计算机软件上实现。 4.攻击密码的两个通用方法是什么？密码分析与穷举法（暴力解码） 5.为什么一些分组密码操作模式只使用了加密，而其他的操作模式使用了加密又使用了解密答：出于加密与解密的考虑，一个密码模式必须保证加密与解密的可逆性。在密码分组链接模式中，对明文与前一密文分组异或后加密，在解密时就要先解密再异或才能恢复出明文；在计数器模式中，对计数器值加密后与明文异或产生密文，在解密时，只需要相同的计数器加密值与密文异或就可得到明文。 6.为什么3DES的中间部分是解密而不是加密？ 3DES加密过程中使用的解密没有密码方面的意义。唯一好处是让3des使用者能解密原来单重des使用者加密的数据。第三章

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